Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To unlock all 5,300 videos, start your free trial.
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
The fundamental theorem of calculus states that if a function f has an antiderivative F, then the definite integral of f from a to b is equal to F(b)-F(a).
Using this, you can solve for the area bounded by a curve and the x-axis. To do so, first find the points at which the graph crosses the x-axis by solving for x when f(x)=0. The results will be the limits of integration. Then, find the antiderivative, and take the integral of the function, and evaluate at the limits of integration solved for previously. Remember that when taking definite integrals, it is possible for the result to be negative (this means that the curve is below the x-axis).
Let’s use the fundamental theorem of calculus to solve another problem. Find the exact area bounded between y equals 6 minus x minus x² and the x axis. Here is my diagram of this region. I want to find the exact area of this region. To do so, I’m going to need to figure out exactly where the curve crosses the x axis, because these are going to be my limits of integration.
I need to set this equation equal to 0. 6 minus x minus x², that will give me the x intercepts at these points. I can factor out the -1, actually I can divide out the -1 and get x² plus x minus 6 equals 0. And then factor this. The factors will be of the form x plus or minus something, x plus or minus something. I’m seeing 2 and 3 here. S plus 3, x minus 2. -3 and 2 are my intercepts. That’s important because it tells me I have to integrate from -3 to 2.
The exact area of this region is going to be the integral from -3 to 2 of 6 minus x minus x²dx. Let’s use the fundamental theory of calculus on this. I need an antiderivative for this function. So I'll put my antiderivative in brackets here. [6x minus 1/2x² minus 1/3x³]. I need to evaluate that from -3 to 2.
First, I evaluate this thing at 2. I get 6 times 2 minus ½ times 4, minus 1/3 times 8, minus, and then I evaluate all of these at -3. 6 times -3 minus ½ times 9 minus 1/3 times -27. Here, I have 12 minus 2 is 10, minus 8/3. 10 minus 8/3, minus, here I have -18 minus 4/2. 27 divided by 3 is 9, so this is going to be +9. So -18 plus 9 is minus 9. Minus 9 minus 9/2.
Let me just simplify this a little bit more. 10 plus 9 minus 8/3 plus 9/2. So 9/2 minus 8/3. 19. I need to get a common denominator to subtract these. The common denominator will be 6, so I multiply the top and bottom by 3. This is 27/6 minus, if you multiply the top and bottom by 2, I’ll get a common denominator. So 16/6, this is 11/6. 19 plus 11/6. Which is the same as 20 and 5/6. That’s my area.
Unit
The Definite Integral